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Brums [2.3K]
3 years ago
12

Name this three dimensional figure​

Mathematics
1 answer:
vodomira [7]3 years ago
8 0

Answer:

Square Pyramid

Step-by-step explanation:

Hope this helps!

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Answer:

5.25 + 2.50 = 7.75. 10 - 7.75 =2.25

Step-by-step explanation:


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David and Lacey are 2000 feet apart on a coastline. They each look at the same boat in the water. The angle between the coastlin
Ulleksa [173]

By using the given distances and directions, we have;

Part A: 1083.5 feet

Part B: 1921.37 feet

Part C: 58.05°

<h3>How can the distances and directions be calculated?</h3>

The distance between David and Lacey = 2000 feet

David's angle to the boat = 70°

Lacey's angle to the boat = 32°

Part A

The distance of the boat from David is found as follows;

Imaginary lines drawn from the boat to David and then to Lacey form a triangle.

In the triangle, let <em>A</em><em> </em>= 70°

B = 32°

Therefore by the sum of angles in a triangle, we have;

C = 180° - (70° + 32°) = 78°

By using sine rule we have;

\frac{a}{sin(A)}  =  \frac{b}{sin( B )} =   \frac{c}{sin(C)}

David's distance from the boat, <em>b</em>, is therefore;

\frac{b}{sin( 32 )} =   \frac{2000}{sin(78)}

The angle subtended by the coastline, <em>C</em>, is therefore;

b =   \frac{2000}{sin(78)}  \times sin( 32 ) = 1083.5

  • David's distance from the boat is 1083.5 feet

Part B

The distance between the boat and Lacey is found as follows;

\mathbf{ \frac{a}{sin( 70)} }=   \frac{2000}{sin(78)}

a =   \frac{2000}{sin(78)}  \times sin( 70 ) = 1921.37

  • Lacey's distance from the boat is 1921.37 feet

Part C

When <em>b </em>= 1200 feet, we have;

Finding the vertical distance of the boat from the coastline, we have;

1083.5 × sin(70) = 1018.17

We have;

\frac{1200}{sin(90)}  =  \frac{1018.17}{sin( B' )}

{sin( B' )} = \frac{sin(90)}{1200} \times 1018.17

The angle between the coastline and his view to the boat, <em>B'</em>, is therefore;

  • B' = arcsine (1018.17÷1200) = 58.05°

Learn more about sine rule here:

brainly.com/question/4372174

#SPJ1

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